Answer
$c(H_{2}SO_{4})=0.0952M$
Work Step by Step
The reaction between sulfuric acid and sodium carbonate can be shown by the following reaction:
$H_{2}SO_{4} + Na_{2}CO_{3} \rightarrow Na_{2}SO_{4} + H_{2}O+CO_{2}$
By observing stoichiometric coefficients, we can conclude that the amount of sulfuric acid required is equal to the amount of sodium carbonate. Hence,
$n(H_{2}SO_{4}) = n(Na_{2}CO_{3}) = \frac{m(Na_{2}CO_{3})}{M(Na_{2}CO_{3})}=\frac{0.3911g}{106\frac{g}{mol}}=0.00369mol$
The molarity of sulfuric acid solution can now be easily calculated:
$c(H_{2}SO_{4})=\frac{n(H_{2}SO_{4})}{V(H_{2}SO_{4})}=\frac{0.00369mol}{0.03875dm^3}=0.0952M$
